# How can we prove that the limit of sin(1/x) does not exist?

• michonamona
In summary, the problem is trying to prove that the limit of sin(1/x) does not exist as x approaches 0. This is done by assuming the limit exists and choosing a small epsilon value, then showing that there is no corresponding delta value for which sin(1/x) will always fall within epsilon of the limit. This contradiction proves that the limit does not exist.
michonamona

## Homework Statement

Prove that the $$lim_{n\rightarrow0}sin(\frac{1}{x}) = DNE$$

## The Attempt at a Solution

In this case we have:

For every epsilon there exists a delta such that:

$$|x|<\delta$$ implies $$|sin(\frac{1}{x})|<\epsilon$$

I know that $$|sin(\frac{1}{x})|\leq1$$. I'm not sure how to proceed after this point.

M

The condition you wrote applies when the limit is zero. If the limit is L, then for every epsilon>0 there exists a delta such that for
|x| < delta we have |sin(x) - L| < epsilon.

Now what do you know about the range of values the function
sin(1/x) will attain for x in the interval 0< |x| < delta for some arbitrary small delta?

Count Iblis said:
The condition you wrote applies when the limit is zero. If the limit is L, then for every epsilon>0 there exists a delta such that for
|x| < delta we have |sin(x) - L| < epsilon.

Now what do you know about the range of values the function
sin(1/x) will attain for x in the interval 0< |x| < delta for some arbitrary small delta?

Ok, I understand that |sin(1/x)| <= 1. So fo some arbitrarily small delta that values of sin(1/x) will be between -1<=sin(1/x)<=1. What can I take away from this?

Thank you

M

michonamona said:
Ok, I understand that |sin(1/x)| <= 1. So fo some arbitrarily small delta that values of sin(1/x) will be between -1<=sin(1/x)<=1. What can I take away from this?

Thank you

M

And all these values between -1 and 1 will actually be attained. The existence of a limit L, however, requires that you can specify some arbitrary small interval of size epsilon containing L and then for all x smaller than some small delta, the function will only attain values in that small interval. If the limit exist then, in general, the smaller you choose epsilon, the smaller the delta will be that you need for this to work. But the point is that you can always find the delta, no matter how small you choose epsilon.

Count Iblis said:
And all these values between -1 and 1 will actually be attained. The existence of a limit L, however, requires that you can specify some arbitrary small interval of size epsilon containing L and then for all x smaller than some small delta, the function will only attain values in that small interval. If the limit exist then, in general, the smaller you choose epsilon, the smaller the delta will be that you need for this to work. But the point is that you can always find the delta, no matter how small you choose epsilon.

In this case, therefore, it won't work because no matter how small we make our delta around x, sin(1/x) will vacillate from 1 to -1. It will never get to a single limit point. Is this correct?

This is intuitive, but the question is how can I translate this mathematically? So going back to my original question, I'm stuck in$$|sin(\frac{1}{x})|\leq1$$ how do I proceed to complete the proof?

I appreciate all of your help,

M

Yes, that's correct. What you do is you assume that the limit exists and is equal to some L. Then you choose epsilon smaller than 1 and show that no value for delta exists such that sin(1/x) for
0<x<delta will always fall within epsilon of L. This thus yields a contradiction.

## 1. What is an Epsilo-Delta proof?

An Epsilo-Delta proof is a method used in mathematics to prove the limit of a function at a specific point. It involves choosing a value for epsilon (ε) and finding a corresponding value for delta (δ) that satisfies a specific condition. This condition is that if the distance between the input value and the specific point (x) is less than delta (|x-a| < δ), then the distance between the output value and the limit (f(x) - L) is less than epsilon (|f(x) - L| < ε).

## 2. What is the significance of using an Epsilo-Delta proof?

An Epsilo-Delta proof is used to rigorously prove the existence of a limit of a function at a specific point. It provides a precise definition of what it means for a function to approach a certain limit, and allows for a clear understanding of the behavior of a function near a specific point.

## 3. How is an Epsilo-Delta proof used to prove the limit of Sin(1/x)?

To prove the limit of Sin(1/x) as x approaches 0 using an Epsilo-Delta proof, we first choose a value for epsilon (ε) and find a corresponding value for delta (δ). Then, we show that for all values of x satisfying |x| < δ, the distance between Sin(1/x) and the limit, which is 0, is less than epsilon (|Sin(1/x) - 0| < ε). This can be done by using trigonometric identities and algebraic manipulation.

## 4. What are some common challenges encountered when using an Epsilo-Delta proof?

One common challenge when using an Epsilo-Delta proof is finding the appropriate values for epsilon and delta. It can also be difficult to manipulate the inequalities and equations involved in the proof to show that they satisfy the condition for convergence. Additionally, understanding the concept of limits and applying the proof correctly can also be challenging for some students.

## 5. How is an Epsilo-Delta proof related to the concept of continuity?

An Epsilo-Delta proof is closely related to the concept of continuity. A function is continuous at a specific point if and only if the limit at that point exists and is equal to the output value at that point. Therefore, an Epsilo-Delta proof can be used to prove the continuity of a function at a specific point by showing that the limit of the function equals the output value at that point.

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